checking polarity without voltmeter

Windows7ge · May 11, 2017

4 minutes ago, bob345 said:

try plating something in an electrolyte solution to determine the polarity. the setup would be something like this using a copper penny and a quarter

Do you think this could be done using 12V on an ATX PSU because with this knowledge I now have the desire to have a copper coated quarter, lol.

themctipers · May 11, 2017

1 minute ago, Windows7ge said:

Do you think this could be done using 12V on an ATX PSU because with this knowledge I now have the desire to have a copper coated quarter, lol.

dont have penny, not in us

nope, it's 15v+ (15-25v, 1a+)

bob345 · May 11, 2017

Just now, Windows7ge said:

Do you think this could be done using 12V on an ATX PSU because with this knowledge I now have the desire to have a copper coated quarter, lol.

Easily, it can even be done with a simple 9 volt battery. Electroplating in a simple electrolyte at a low voltage is not a very power hungry process.

bob345 · May 11, 2017

3 minutes ago, themctipers said:

dont have penny, not in us

nope, it's 15v+ (15-25v, 1a+)

You could technically use the copper wire itself as the cathode as long as it is not tinned. In that case any other kind of coin or piece of steel can be used.

Windows7ge · May 11, 2017

9 minutes ago, bob345 said:

Easily, it can even be done with a simple 9 volt battery. Electroplating in a simple electrolyte at a low voltage is not a very power hungry process.

Awesome. Now I just need a penny made before 1982.

mariushm · May 11, 2017

Find a diode and a thin wire . You can take a diode out of any old or broken thing that you have around.

If you put a thin wire across the wires of your power supply it will overheat and break (if it's thin enough). So that will tell you if there's flow of current or not.

A diode lets current flow only in one direction so if you put it one way it lets electricity flow, if you put it the wrong way it won't.

[+ psu ] ============ [A ==>| C ] =========[wire end -------- wire end ] ============[ - psu ]

a = anode of diode , c = cathode of diode , the part with the vertical bar on the component.

If you make this circuit and the wire breaks, then the wire on the left side of the diode (the one without the vertical bar) is the POSITIVE (+19v).

If the wire doesn't heat up, the wire on the left side is the negative and the diode blocks the electricity flow.

A LED is also a diode which produces light in only one direction, but the problem is they can only block flow if the voltage is below around 5v . So you can't test with a single led unless you drop some voltage on a resistor or unless you put multiple leds in series, so that each one would see less than 5v if you put the wires the wrong way

A red led will need about 2v to produce light, green leds about 2.2v and white/blue leds need about 3..3.2v to light up ... let's go with 2v

For a 19v power supply, you could use 10 such leds in series without risking them being damaged if connected the wrong way, because each led will see about 1.8..2v across them

[+ psu] ===== [+ led -]=[+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= ==== [-] psu

leds light up = left side is positive ,

leds don't light up = left side is negative

for a single led, you calculate how many volts may drop on the resistor at a particular test current , let's say 10mA (0.01A)

V=IxR ... 19v - 2v (forward voltage of led) = 0.01 x R == > R = 17/0.01 = 1700 ohm .. so use 1500-2500 ohm to limit the current to around 5-15 mA (most leds can tolerate easily up to 20mA) .. i'd just go with two 1000 ohm resistors in series, or a 1500 ohm or a 1800 ohm or a 2200 ohm resistor, all standard values and easy to find.

[+ psu ] ====== [ resistor ] = [+ led -] ==== [- psu]

led lights up = left side is positive

led doesn't light up = left side is negative wire.

Another option, though ever so slightly a bit risky would be to get a regular computer fan and use two resistors in a voltage divider arrangement and connect the fan .. the voltage divider reduces the voltage the fan will see to around 5v to 10v so if you connect it for just a brief moment if you connect it the right way it should try to spin in the right direction. If the polarity is wrong, the fan won't move. The risky part is that some fans don't handle polarity well and even with current limited by the resistors in the voltage divider, the fan may still be damaged.

[+ psu ] ======== [ resistor ] ===== [ * to + wire of fan * ] ===== [ resistor ] ======[ - psu ] ===== [ to - wire of fan ]

resistors should be in the 4.7k .. 10k range

themctipers · May 11, 2017

57 minutes ago, mariushm said:

Find a diode and a thin wire . You can take a diode out of any old or broken thing that you have around.

If you put a thin wire across the wires of your power supply it will overheat and break (if it's thin enough). So that will tell you if there's flow of current or not.

A diode lets current flow only in one direction so if you put it one way it lets electricity flow, if you put it the wrong way it won't.

[+ psu ] ============ [A ==>| C ] =========[wire end -------- wire end ] ============[ - psu ]

a = anode of diode , c = cathode of diode , the part with the vertical bar on the component.

If you make this circuit and the wire breaks, then the wire on the left side of the diode (the one without the vertical bar) is the POSITIVE (+19v).

If the wire doesn't heat up, the wire on the left side is the negative and the diode blocks the electricity flow.

A LED is also a diode which produces light in only one direction, but the problem is they can only block flow if the voltage is below around 5v . So you can't test with a single led unless you drop some voltage on a resistor or unless you put multiple leds in series, so that each one would see less than 5v if you put the wires the wrong way

A red led will need about 2v to produce light, green leds about 2.2v and white/blue leds need about 3..3.2v to light up ... let's go with 2v

For a 19v power supply, you could use 10 such leds in series without risking them being damaged if connected the wrong way, because each led will see about 1.8..2v across them

[+ psu] ===== [+ led -]=[+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= [+ led -]= ==== [-] psu

leds light up = left side is positive ,

leds don't light up = left side is negative

for a single led, you calculate how many volts may drop on the resistor at a particular test current , let's say 10mA (0.01A)

V=IxR ... 19v - 2v (forward voltage of led) = 0.01 x R == > R = 17/0.01 = 1700 ohm .. so use 1500-2500 ohm to limit the current to around 5-15 mA (most leds can tolerate easily up to 20mA) .. i'd just go with two 1000 ohm resistors in series, or a 1500 ohm or a 1800 ohm or a 2200 ohm resistor, all standard values and easy to find.

[+ psu ] ====== [ resistor ] = [+ led -] ==== [- psu]

led lights up = left side is positive

led doesn't light up = left side is negative wire.

Another option, though ever so slightly a bit risky would be to get a regular computer fan and use two resistors in a voltage divider arrangement and connect the fan .. the voltage divider reduces the voltage the fan will see to around 5v to 10v so if you connect it for just a brief moment if you connect it the right way it should try to spin in the right direction. If the polarity is wrong, the fan won't move. The risky part is that some fans don't handle polarity well and even with current limited by the resistors in the voltage divider, the fan may still be damaged.

[+ psu ] ======== [ resistor ] ===== [ * to + wire of fan * ] ===== [ resistor ] ======[ - psu ] ===== [ to - wire of fan ]

resistors should be in the 4.7k .. 10k range